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If 
alpha,beta,gamma,delta in R satisfy 
((alpha+1)^(2)+(beta+1)^(2)+(gamma+1)^(2)+(delta+1)^(2))/(alpha+beta+gamma+delta)=4 If biquadratic equation 
a_(0)x^(4)+a_(1)x^(3)+a_(2)x^(2)+a_(3)x+a_(4)=0 has the roots 
(alpha+(1)/(beta)-1),(beta+(1)/(gamma)-1),(gamma+(1)/(delta)-1),(delta+(1)/(alpha)-1). Then the value of 
a_(2)//a_(0) is :
(a) 4
(b) -4
(c) 6
(d) none of these

If α,β,γ,δR\alpha,\beta,\gamma,\delta \in \mathbb{R} satisfy (α+1)2+(β+1)2+(γ+1)2+(δ+1)2α+β+γ+δ=4\frac{(\alpha+1)^{2}+(\beta+1)^{2}+(\gamma+1)^{2}+(\delta+1)^{2}}{\alpha+\beta+\gamma+\delta}=4 If biquadratic equation a0x4+a1x3+a2x2+a3x+a4=0a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}=0 has the roots (α+1β1),(β+1γ1),(γ+1δ1),(δ+1α1)(\alpha+\frac{1}{\beta}-1),(\beta+\frac{1}{\gamma}-1),(\gamma+\frac{1}{\delta}-1),(\delta+\frac{1}{\alpha}-1). Then the value of a2a0\frac{a_{2}}{a_{0}} is :\newline(a) 44\newline(b) 4-4\newline(c) 66\newline(d) none of these

Full solution

Q. If α,β,γ,δR\alpha,\beta,\gamma,\delta \in \mathbb{R} satisfy (α+1)2+(β+1)2+(γ+1)2+(δ+1)2α+β+γ+δ=4\frac{(\alpha+1)^{2}+(\beta+1)^{2}+(\gamma+1)^{2}+(\delta+1)^{2}}{\alpha+\beta+\gamma+\delta}=4 If biquadratic equation a0x4+a1x3+a2x2+a3x+a4=0a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}=0 has the roots (α+1β1),(β+1γ1),(γ+1δ1),(δ+1α1)(\alpha+\frac{1}{\beta}-1),(\beta+\frac{1}{\gamma}-1),(\gamma+\frac{1}{\delta}-1),(\delta+\frac{1}{\alpha}-1). Then the value of a2a0\frac{a_{2}}{a_{0}} is :\newline(a) 44\newline(b) 4-4\newline(c) 66\newline(d) none of these
  1. Expand and Simplify: Given the equation:\newline(α+1)2+(β+1)2+(γ+1)2+(δ+1)2α+β+γ+δ=4\frac{(\alpha+1)^{2}+(\beta+1)^{2}+(\gamma+1)^{2}+(\delta+1)^{2}}{\alpha+\beta+\gamma+\delta}=4\newlineWe can simplify this equation by expanding the squares and then dividing by the sum of α\alpha, β\beta, γ\gamma, and δ\delta.
  2. Divide by Sum: Expanding the numerator, we get: \newline(α2+2α+1)+(β2+2β+1)+(γ2+2γ+1)+(δ2+2δ+1)(\alpha^2 + 2\alpha + 1) + (\beta^2 + 2\beta + 1) + (\gamma^2 + 2\gamma + 1) + (\delta^2 + 2\delta + 1)\newlineThis simplifies to:\newline(α2+β2+γ2+δ2)+2(α+β+γ+δ)+4(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) + 2(\alpha + \beta + \gamma + \delta) + 4
  3. Set up Equation: Now, we divide by the sum of α\alpha, β\beta, γ\gamma, and δ\delta:
    (α2+β2+γ2+δ2)+2(α+β+γ+δ)+4α+β+γ+δ\frac{(\alpha^2 + \beta^2 + \gamma^2 + \delta^2) + 2(\alpha + \beta + \gamma + \delta) + 4}{\alpha + \beta + \gamma + \delta}
    This simplifies to:
    (α+β+γ+δ)+2+4α+β+γ+δ(\alpha + \beta + \gamma + \delta) + 2 + \frac{4}{\alpha + \beta + \gamma + \delta}
    Given that the whole expression equals 44, we can set up the equation:
    (α+β+γ+δ)+2+4α+β+γ+δ=4(\alpha + \beta + \gamma + \delta) + 2 + \frac{4}{\alpha + \beta + \gamma + \delta} = 4
  4. Solve for S: We can now solve for (α+β+γ+δ)(\alpha + \beta + \gamma + \delta):(α+β+γ+δ)+4(α+β+γ+δ)=2(\alpha + \beta + \gamma + \delta) + \frac{4}{(\alpha + \beta + \gamma + \delta)} = 2Let's denote S=α+β+γ+δS = \alpha + \beta + \gamma + \delta. Then we have:S+4S=2S + \frac{4}{S} = 2Multiplying both sides by SS to clear the fraction, we get:S2+4=2SS^2 + 4 = 2S
  5. Quadratic Equation: Rearranging the terms, we get a quadratic equation in SS: \newlineS22S+4=0S^2 - 2S + 4 = 0\newlineThis equation does not have real roots, which means there is a math error in our previous steps. We need to re-evaluate our approach to solving for SS.

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